The logarithm is in BMO but not in L-infinity

Local integrability and the scaling reduction. For every $R<\mathrm{\infty }$ the shell bound of [L2] gives ${\int }_{\mathrm{\mid }z\mathrm{\mid }\le R}\mathrm{\mid }\log \mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }dz<\mathrm{\infty }$, because $\{\mathrm{\mid }z\mathrm{\mid }\le 1\}$ is covered by the shells $j\ge 0$ with total ${\sum }_j{2}^{(1-j)n}(j+1)\log 2<\mathrm{\infty }$, while on $\{1<\mathrm{\mid }z\mathrm{\mid }\le R\}$ one has $\mathrm{\mid }\log \mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }\le \log R$ and the enclosing cube $[-R,R{]}^n$ has volume $(2R{)}^n$; hence $b\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^n)$. For a cube $Q$ write $x=x_Q+ry$ with $y\in [-\frac{1}{2},\frac{1}{2}{]}^n$, so that $\mathrm{\mid }Q\mathrm{\mid }=r^n$ and, by [L2], $\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_Q\mathrm{\mid }b-c\mathrm{\mid }={\int }_{[-\frac{1}{2},\frac{1}{2}{]}^n}\mid \log r+\log \mathrm{\mid }\xi +y\mathrm{\mid }-c\mid dy$ with $\xi :=x_Q\mathrm{/}r$; choosing $c=\log r+c_{\xi }$ shows that it suffices to bound $A(\xi ):={\int }_{[-\frac{1}{2},\frac{1}{2}{]}^n}\mathrm{\mid }\log \mathrm{\mid }\xi +y\mathrm{\mid }-c_{\xi }\mathrm{\mid }dy$ uniformly in $\xi$, and then [L1] gives $\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_Q\mathrm{\mid }b-b_Q\mathrm{\mid }\le 2A(\xi )$.

The regular case $\mathrm{\mid }\xi \mathrm{\mid }>2\sqrt{n}$. With $c_{\xi }=\log \mathrm{\mid }\xi \mathrm{\mid }$, for $y\in [-\frac{1}{2},\frac{1}{2}{]}^n$ and $t\in [0,1]$ one has $\mathrm{\mid }\xi +ty\mathrm{\mid }\ge \mathrm{\mid }\xi \mathrm{\mid }-\frac{1}{2}\sqrt{n}>\frac{3}{2}\sqrt{n}$, so $\log \mathrm{\mid }z\mathrm{\mid }$ is differentiable along the segment and $\mid \log \mathrm{\mid }\xi +y\mathrm{\mid }-\log \mathrm{\mid }\xi \mathrm{\mid }\mid \le {\int }_0^1\frac{\mathrm{\mid }y\mathrm{\mid }}{\mathrm{\mid }\xi +ty\mathrm{\mid }}dt\le \frac{\sqrt{n}\mathrm{/}2}{3\sqrt{n}\mathrm{/}2}=\frac{1}{3}$; hence $A(\xi )\le \frac{1}{3}$.

The singular case $\mathrm{\mid }\xi \mathrm{\mid }\le 2\sqrt{n}$. With $c_{\xi }=0$ the shifted cube lies in $\{\mathrm{\mid }z\mathrm{\mid }\le 2\sqrt{n}+\frac{1}{2}\sqrt{n}\}\subseteq \{\mathrm{\mid }z\mathrm{\mid }\le 3\sqrt{n}\}$, so $A(\xi )\le {\int }_{\mathrm{\mid }z\mathrm{\mid }\le 3\sqrt{n}}\mathrm{\mid }\log \mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }dz\le C_n$ by the estimates of step 1.1 with $R=3\sqrt{n}$.

Combining steps 2.1 and 1.2, ${\sup }_{\xi }A(\xi )\le C_n<\mathrm{\infty }$, so [L1] and step 1.1 give $\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}\le 2C_n$: the seminorm depends only on $n$ and $b\in \mathrm{B}\mathrm{M}\mathrm{O}({\mathbb{R}}^n)$.

Unboundedness. As $x\to 0$ one has $\log \mathrm{\mid }x\mathrm{\mid }\to -\mathrm{\infty }$, so $b$ is unbounded on every ball $B(0,\epsilon )$, hence on every neighbourhood of $0$. If $g\in L^{\mathrm{\infty }}({\mathbb{R}}^n)$ satisfied $g=b$ almost everywhere, then for every $M>\mathrm{\parallel }g{\mathrm{\parallel }}_{L^{\mathrm{\infty }}}$ the set $\{x:\mathrm{\mid }b(x)\mathrm{\mid }>M\}$ would have positive measure (it contains $B(0,\epsilon )\setminus \{0\}$ for $\epsilon =e^{-M}$, which has positive measure because the ball contains a nondegenerate cube and a singleton has measure zero) while $\{\mathrm{\mid }g\mathrm{\mid }>M\}$ is null, contradicting almost-everywhere equality; so no bounded function represents the class of $b$. If the class of $b$ were the image of $[g]$ with $g\in L^{\mathrm{\infty }}$, then $b=g+c$ almost everywhere for a constant $c$, and $g+c$ would be a bounded representative, which is impossible. Thus the class of $b$ is not in the image of $L^{\mathrm{\infty }}\mathrm{/}\mathbb{C}$, and the continuous injection is not surjective.